Calculus of Real and Complex Variables

2.4 Root Test

The root test has to do with when a series of complex numbers converges. I am assuming the reader has
been exposed to infinite series. However, this that I am about to explain is a little more general than what
is usually seen in calculus.

Theorem 2.4.1Let ak∈ Fpand consider∑k=1∞ak. Then this series converges absolutelyif

lim sup |ak|1∕k = r < 1.
k→∞

The series diverges spectacularly if limsupk→∞

|ak|

1∕k> 1 and iflimsupk→∞

|ak|

1∕k = 1, the testfails.

Proof:Suppose first that limsupk→∞

|ak|

1∕k = r < 1. Then letting R ∈

(r,1)

, it follows from the
definition of limsup that for all k large enough,

|ak|1∕k ≤ R

Hence there exists N such that if k ≥ N, then

|a |
k

≤ Rk. Let Mk =

|a |
k

for k < N and let Mk = Rk for
k ≥ N. Then

∞∑ N∑−1 N
Mk ≤ |ak|+ -R--- < ∞
k=1 k=1 1 − R

and so, by the Weierstrass M test applied to the series of constants, the series converges and also converges
absolutely. If

lim sup |ak|1∕k = r > 1,
k→∞

then letting r > R > 1, it follows that for infinitely many k,

k
|ak| > R

and so there is a subsequence which is unbounded. In particular, the series cannot converge and in fact
diverges spectacularly. In case that the limsup = 1, you can consider ∑n=1∞

1
n

which diverges by calculus
and ∑n=1∞

1
n2

which converges, also from calculus. However, the limsup equals 1 for both of these.
■

This is a major theorem because the limsup always exists. As an important application, here is a
corollary.

Corollary 2.4.2If∑kakconverges, then limsupk→∞

|ak|

1∕k≤ 1.

If the sequence has values in X a complete normed linear space, there is no change in the
conclusion or proof of the above theorem. You just replace